Power Series Proof Workshop

A Local Question About Reciprocals

The Cauchy Product Theorem explains when two absolutely convergent power series can be multiplied. A natural next question is whether a power series can also be divided into another one. The essential local case is the reciprocal: if a power series has a nonzero value at its center, can its reciprocal be represented by a power series near that center?

The proof uses a useful strategy. Separate the constant term from the rest, then shrink the interval until the remaining part is smaller in absolute value than the constant term. A geometric series can then construct the reciprocal. The main work is to justify that the resulting coefficients form a convergent power series and that grouping them gives the desired function.

Let

A(x)=\sum_{n=0}^{\infty}a_n(x-a)^n

have positive radius of convergence, and suppose $a_0\neq0$. Write $H(x)=\sum_{n=1}^{\infty}a_n(x-a)^n$, so $A(x)=a_0+H(x)$. We first establish that the nonconstant part can be made small in absolute value.

Lemma (Small-Radius Tail Bound): Suppose $A(x)=\sum_{n=0}^{\infty}a_n(x-a)^n$ has positive radius of convergence. For every $\varepsilon>0$, there is an $r>0$ such that $$ \sum_{n=1}^{\infty}|a_n|r^n<\varepsilon. $$

Proof. Choose $s>0$ strictly smaller than the radius of convergence. By Convergence Inside and Outside the Radius, the series $\sum_{n=1}^{\infty}|a_n|s^n$ converges; call its sum $K$. For $0<r\leq s$ and every $n\geq1$, $(r/s)^n\leq r/s$. Therefore

\sum_{n=1}^{\infty}|a_n|r^n =\sum_{n=1}^{\infty}|a_n|s^n(r/s)^n \leq\frac{r}{s}\sum_{n=1}^{\infty}|a_n|s^n =\frac{rK}{s}.

If $K=0$, the sum on the left is zero for every $0<r\leq s$. If $K>0$, choose $r$ with $0<r\leq s$ and $rK/s<\varepsilon$. In either case such an $r$ exists, and the claimed inequality follows. $\square$

Constructing the Coefficients

Apply the lemma with $\varepsilon=|a_0|$. We can choose $r>0$, still smaller than the original radius of convergence, so that

h(r):=\sum_{n=1}^{\infty}|a_n|r^n<|a_0|.

For a fixed $x$ with $|x-a|\leq r$, the series for $H(x)$ converges absolutely and $|H(x)|\leq h(r)<|a_0|$. Thus the scalar geometric series suggests

\frac{1}{A(x)} =\frac{1}{a_0+H(x)} =\frac{1}{a_0}\sum_{k=0}^{\infty} \left(-\frac{H(x)}{a_0}\right)^k.

To turn this expression into a power series in $x-a$, expand each power of $H(x)$ using the Cauchy Product Theorem. For $n\geq1$, define the finite sum

q_n=\sum_{k=1}^{n}\frac{(-1)^k}{a_0^{k+1}} \sum_{\substack{n_1,\ldots,n_k\geq1\\n_1+\cdots+n_k=n}} a_{n_1}\cdots a_{n_k}, \qquad q_0=\frac1{a_0}.

The inner sum is over the ways to write $n$ as a sum of $k$ positive integers. It is finite, and there are no such choices when $k>n$. Consequently each $q_n$ is a well-defined finite expression in the original coefficients. This finiteness is important: it allows us to define the coefficients before addressing the convergence of the resulting infinite series.

Theorem (Local Power-Series Reciprocal): Suppose $A(x)=\sum_{n=0}^{\infty}a_n(x-a)^n$ has positive radius of convergence and $a_0\neq0$. There is an $r>0$ such that the coefficients $q_n$ defined above satisfy $$ \sum_{n=0}^{\infty}|q_n|r^n<\infty, \qquad \frac{1}{A(x)}=\sum_{n=0}^{\infty}q_n(x-a)^n \quad\text{for }|x-a|\leq r. $$

Proof. Choose $r$ so that $h(r)<|a_0|$, as above. For a fixed $k\geq1$, the absolute values of all products arising from $k$ factors of $H$, weighted by their powers of $r$, have total

\sum_{n_1,\ldots,n_k\geq1} |a_{n_1}|\cdots|a_{n_k}|r^{n_1+\cdots+n_k} =\left(\sum_{m=1}^{\infty}|a_m|r^m\right)^k =h(r)^k.

This equality follows by multiplying finite partial sums and then letting their bounds increase; all terms are nonnegative. Using the triangle inequality in the definition of $q_n$, and summing these bounds over $k$, gives

\sum_{n=0}^{\infty}|q_n|r^n \leq \frac1{|a_0|} +\sum_{k=1}^{\infty}\frac{h(r)^k}{|a_0|^{k+1}} =\frac1{|a_0|}\sum_{k=0}^{\infty} \left(\frac{h(r)}{|a_0|}\right)^k <\infty.

The final series is geometric with ratio strictly less than $1$. The comparison proves absolute convergence of the reciprocal power series at every $x$ with $|x-a|\leq r$.

It remains to identify its sum. For each such $x$, the absolute sum over all choices of $k$ and $n_1,\ldots,n_k$ is bounded by the same convergent geometric series just displayed. Thus the terms can be grouped by their total index $n=n_1+\cdots+n_k$; each group gives exactly $q_n(x-a)^n$. The scalar geometric-series formula therefore gives

\sum_{n=0}^{\infty}q_n(x-a)^n =\frac1{a_0}\sum_{k=0}^{\infty} \left(-\frac{H(x)}{a_0}\right)^k =\frac1{a_0+H(x)} =\frac1{A(x)}.

Here $|H(x)/a_0|\leq h(r)/|a_0|<1$, so the geometric series is valid and $a_0+H(x)\neq0$. This proves both the convergence and the reciprocal identity. $\square$

Worked Applications

Worked Example: The Reciprocal of a Quadratic

Consider $A(x)=2+x+x^2$, centered at $a=0$. At $r=1/2$, the absolute sum of the nonconstant terms is

h(1/2)=\frac12+\frac14=\frac34<2=|a_0|.

The theorem guarantees a convergent power series for $1/(2+x+x^2)$ when $|x|\leq1/2$. The coefficients can also be computed directly by requiring the product with $2+x+x^2$ to have constant coefficient $1$ and all higher coefficients $0$. The first equations are

2q_0=1,\qquad 2q_1+q_0=0,\qquad 2q_2+q_1+q_0=0,\qquad 2q_3+q_2+q_1=0.

Solving in order gives $q_0=1/2$, $q_1=-1/4$, $q_2=-1/8$, and $q_3=3/16$. For instance, the coefficient at index $2$ checks because $2(-1/8)+(-1/4)+(1/2)=0$, and at index $3$ because $2(3/16)+(-1/8)+(-1/4)=0$. Hence

\frac1{2+x+x^2} =\frac12-\frac{x}{4}-\frac{x^2}{8}+\frac{3x^3}{16}+\cdots \qquad (|x|\leq1/2).

The theorem supplies convergence and equality for the full series; the coefficient equations alone would only provide a formal list of coefficients.

Worked Example: A Reciprocal with a Vanishing Quadratic Coefficient

Let $A(x)=1+x+x^2$. Choose $r=1/3$. Then

h(1/3)=\frac13+\frac19=\frac49<1=|a_0|.

So the reciprocal has a convergent power series on $|x|\leq1/3$. Writing $Q(x)=\sum_{n=0}^{\infty}q_nx^n$, the product identity $A(x)Q(x)=1$ yields $q_0=1$, $q_1=-1$, $q_2=0$, $q_3=1$, and $q_4=-1$. The checks at the first indices are

q_0=1,\quad q_1+q_0=-1+1=0,\quad q_2+q_1+q_0=0-1+1=0,\quad q_3+q_2+q_1=1+0-1=0.

Thus the beginning of the expansion is $1-x+0x^2+x^3-x^4+\cdots$. The zero coefficient is not an error: it records cancellation among the terms contributing to that index. The convergence conclusion comes from the reciprocal theorem, not from observing a pattern in the first few coefficients.

Worked Example: The Reciprocal of One Plus the Exponential Series

Let $E(x)=\sum_{n=0}^{\infty}x^n/n!$, the exponential series, and set $A(x)=1+E(x)$. Its constant term is $2$, and its nonconstant coefficients are $a_n=1/n!$ for $n\geq1$. At $r=1/2$,

h(1/2)=\sum_{n=1}^{\infty}\frac{(1/2)^n}{n!} <\sum_{n=1}^{\infty}(1/2)^n=1<2.

The strict inequality holds because $1/n!\leq1$ for every $n\geq1$, with strict inequality for $n\geq2$. The reciprocal theorem therefore gives an absolutely convergent expansion on $|x|\leq1/2$. Computing its first coefficients from the product with $A(x)=2+x+x^2/2!+x^3/3!+\cdots$ gives

q_0=\frac12,\qquad q_1=-\frac14,\qquad q_2=0,\qquad q_3=\frac1{48}.

For the index $2$, the coefficient equation is $2q_2+q_1+q_0/2=0$, and indeed $0-1/4+1/4=0$. For index $3$, it is $2q_3+q_2+q_1/2+q_0/6=0$; substitution gives $1/24+0-1/8+1/12=0$. Consequently

\frac1{1+E(x)} =\frac12-\frac{x}{4}+\frac{x^3}{48}+\cdots \qquad (|x|\leq1/2).

This example illustrates how a known power series can be used as input to a new local expansion, even when the reciprocal coefficients are not immediately recognizable.

What the Proof Does—and Does Not—Say

The crucial hypothesis is $a_0\neq0$, which says that $A(a)\neq0$. If $a_0=0$, then $A(a)=0$, so a reciprocal cannot be defined at the center. For example, $A(x)=x$ has a perfectly valid power series at $0$, but $1/A(x)=1/x$ is not defined there and cannot equal a power series on a neighborhood containing $0$.

A second point is that the coefficient formula involves only finitely many contributions at each index, but the proof of convergence is still necessary. A formal multiplication that produces the desired coefficients does not by itself show that the series converges or that its sum is the reciprocal. Here, the small-radius bound supplies absolute convergence, and the geometric-series identity identifies the sum.

The method is a practical proof pattern: isolate a nonzero constant, bound the remaining series by absolute values on a smaller interval, and then use a geometric expansion whose terms can be regrouped safely. The Cauchy Product Theorem justifies the finite-power expansions, while the absolute bound controls the infinite collection of resulting terms.

Takeaway: A power series with nonzero constant term has a reciprocal represented by a convergent power series on some smaller neighborhood of its center. The nonzero constant term makes the geometric construction possible; absolute convergence makes its coefficient grouping legitimate.

Check Your Understanding

Use the coefficient construction and the convergence argument to answer the following questions.

Why can the sum of the absolute values of the nonconstant terms be made smaller than $|a_0|$ by shrinking the radius?
Why does the formula for $q_n$ contain no terms with $k>n$?
Which estimate proves that the reciprocal coefficients define an absolutely convergent power series?
For $A(x)=3+x$, compute the first three coefficients of its reciprocal expansion at $0$.
Why does a nonzero constant term matter for representing the reciprocal by a power series centered at the same point?

What You'll Learn